Sprawdź następujące tożsamości.

[DaViDTS]

j/w

1) \(\displaystyle{ \frac{1+\cos\alpha}{\sin\alpha}= \cot \frac{a}{2}}\)
2) \(\displaystyle{ \frac{1-\cos\alpha}{\sin\alpha}= \tan \frac{a}{2}}\)

dziekuję.

[Mithrandir]

1) \(\displaystyle{ \frac{1+\cos\alpha}{\sin\alpha}= \frac{1+\cos\ (2*\frac{\alpha}{2}) }{\sin\ (2*\frac{\alpha}{2})}= \frac{1+\cos^{2} \frac{\alpha}{2}-\sin^{2} \frac{\alpha}{2} }{2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}= \frac{1+\cos^{2} \frac{\alpha}{2}-1+\cos^{2} \frac{\alpha}{2} }{2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}= \frac{2\cos^{2}\frac{\alpha}{2}}{2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}=\frac{\cos \frac{\alpha}{2}}{\sin\frac{\alpha}{2}}=ctg\frac{\alpha}{2}}\)

2) Drugie analogicznie.

[Mersenne]

2) \(\displaystyle{ \frac{1-\cos }{\sin }=\tan \frac{\alpha}{2}}\)

zał.: \(\displaystyle{ \begin{cases} \sin \neq 0 \\ \frac{\alpha}{2}\neq \frac{\pi}{2}+k\pi \end{cases} \iff \begin{cases} \neq k\pi \\ \neq \pi+2k\pi \end{cases} \iff \neq k\pi, k\in C}\)

\(\displaystyle{ L=\frac{1-\cos }{\sin }=\frac{\cos 0-\cos }{\sin }=\frac{-2\sin \frac{\alpha}{2}\sin ft(-\frac{\alpha}{2}\right)}{\sin }=\frac{2\sin \frac{\alpha}{2}\sin \frac{\alpha}{2}}{\sin }=\frac{2\sin \frac{\alpha}{2}\sin \frac{\alpha}{2}}{2\sin \frac{\alpha}{2}\cos \frac{\alpha}{2}}=}\)

\(\displaystyle{ =\frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}=\tan \frac{\alpha}{2}}\)

\(\displaystyle{ L=P}\)

c.n.d.